Let $f_A$ be the map from line $BC$ to the plane that projects the input point $P$ onto sides $AB$ and $AC$, and outputs their midpoint. This is clearly a linear function, so the range of $f_A$ is actually a line. We similarly define $f_B$ and $f_C$. Note that $f_B(A)=f_C(A)$ is the midpoint $X$ of $AD$, where $DEF$ is the orthic triangle. Thus, the functions $f_A$, $f_B$, $f_C$ agree where their boundaries meet, so we can actually consolidate them all into one function $f$ from the plane to itself that sends $A$, $B$, $C$ to the midpoints of the altitudes $X$, $Y$, $Z$, and sends \[f(xA+yB+zC)=xf(A)+yf(B)+zf(C)\]for any $x+y+z=1$. (a) We see that $A'=f(A_1)$, $B'=f(A_2)$, $C'=f(C_2)$, and since $f$ is an affine transformation that preserves collinearity, we are done. (b) It suffices to show that $f(O)=N$, or that the barycentric coordinates of $O$ with respect to $ABC$ are the same as those of $N$ with respect to $XYZ$. This amounts to showing that $[NYZ]/[OBC]$ is equal to its two cyclic variations by the areal definition of barycentric coordinates. To make things easier, let $X'$ be the reflection of $H$ over $X$, and define $Y'$ and $Z'$ similarly. Since $N$ is the midpoint of $OH$, it suffices to show that the areas $[OY'Z']/[OBC]$ are constant. We will show this using complex numbers, with $(ABC)$ being the unit circle. Note that \[x'=a+d-h=\frac{a-b-c}{2}-\frac{bc}{2a},\]so defining $x''=-2x$ (and similarly $y''$ and $z''$), it suffices to show that $[OY''Z'']/[OBC]$ is the same for all three sides. Note that \[k:=\frac{[OY''Z'']}{[OBC]}=\frac{y''\bar{z}''-\bar{y}''z''}{b\bar{c}-c\bar{b}}.\]Note that \[y''\bar{z}''-\bar{y}''z''=(a+c-b+ac/b)(1/a+1/b-1/c+c/ab)-(1/a+1/c-1/b+b/ac)(a+b-c+ab/c),\]which can be expanded and factored into \[-\frac{(a+b)(a+c)(b-c)(b+c)^2}{ab^2c^2}.\]Thus, \[k=-\frac{\frac{(a+b)(a+c)(b-c)(b+c)^2}{ab^2c^2}}{\frac{(b-c)(b+c)}{bc}}=-\frac{(a+b)(b+c)(c+a)}{abc},\]which is symmetric in $a,b,c$, as desired. This completes the proof.